RS Aggarwal MCQs: Linear Equations in Two Variables | Mathematics (Maths) Class 9 PDF Download

 Page 1


         
     
     
          
          
          
                  
                        
     
Q u e s t i o n : 2 0
The equation of the x-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the y-coordinate of any point on x-axis is always 0.
So, the equation of the x-axis is y = 0.
Hence, the correct option is b
.
Q u e s t i o n : 2 1
The equation of the y-axis is
a
  x = 0
Page 2


         
     
     
          
          
          
                  
                        
     
Q u e s t i o n : 2 0
The equation of the x-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the y-coordinate of any point on x-axis is always 0.
So, the equation of the x-axis is y = 0.
Hence, the correct option is b
.
Q u e s t i o n : 2 1
The equation of the y-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the x-coordinate of any point on y-axis is always 0.
So, the equation of the y-axis is x = 0.
Hence, the correct option is a
.
Q u e s t i o n : 2 2
The point of the form (a, a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
c
the line y = x
Given, a point of the form (a, a)
, where a
? 0 .
When a = 1
, the point is 1, 1
When a = 2
, the point is 2, 2
.....and so on.
Plot the points 1, 1
and 2, 2
.......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x
.
Q u e s t i o n : 2 3
The point of the form (a, -a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
d
the line x + y = 0
Given, a point of the form (a, -a)
, where a
 ? 0.
When a = 1
, the point is 1, -1
.
When a = 2
, the point is 2, -2
.
When a = 3
, the point is 3, -3
.......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
Page 3


         
     
     
          
          
          
                  
                        
     
Q u e s t i o n : 2 0
The equation of the x-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the y-coordinate of any point on x-axis is always 0.
So, the equation of the x-axis is y = 0.
Hence, the correct option is b
.
Q u e s t i o n : 2 1
The equation of the y-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the x-coordinate of any point on y-axis is always 0.
So, the equation of the y-axis is x = 0.
Hence, the correct option is a
.
Q u e s t i o n : 2 2
The point of the form (a, a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
c
the line y = x
Given, a point of the form (a, a)
, where a
? 0 .
When a = 1
, the point is 1, 1
When a = 2
, the point is 2, 2
.....and so on.
Plot the points 1, 1
and 2, 2
.......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x
.
Q u e s t i o n : 2 3
The point of the form (a, -a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
d
the line x + y = 0
Given, a point of the form (a, -a)
, where a
 ? 0.
When a = 1
, the point is 1, -1
.
When a = 2
, the point is 2, -2
.
When a = 3
, the point is 3, -3
.......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
You will get the equation of the line x + y = 0
.
Q u e s t i o n : 2 4
The linear equation 3x - 5y = has
a a unique solution
b two solutions
c infinitely many solutions
d no solution
S o l u t i o n :
c infinitely many solutions
Given linear equation: 3x - 5y = 15
Or, x = 
5y + 15
3
When y = 0
, x = 
15
3
 = 5
.
When  y = 3
, x = 
30
3
 = 10
.
When y = -3
, x = 
0
3
 = 0
.
Thus, we have the following table:
     x      5      10    0
     y      0      3    -3
Plot the points A(5, 0) , B(10, 3)
 and C(0, -3)
. Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation. 
Hence, the linear equation has infinitely many solutions.
Page 4


         
     
     
          
          
          
                  
                        
     
Q u e s t i o n : 2 0
The equation of the x-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the y-coordinate of any point on x-axis is always 0.
So, the equation of the x-axis is y = 0.
Hence, the correct option is b
.
Q u e s t i o n : 2 1
The equation of the y-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the x-coordinate of any point on y-axis is always 0.
So, the equation of the y-axis is x = 0.
Hence, the correct option is a
.
Q u e s t i o n : 2 2
The point of the form (a, a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
c
the line y = x
Given, a point of the form (a, a)
, where a
? 0 .
When a = 1
, the point is 1, 1
When a = 2
, the point is 2, 2
.....and so on.
Plot the points 1, 1
and 2, 2
.......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x
.
Q u e s t i o n : 2 3
The point of the form (a, -a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
d
the line x + y = 0
Given, a point of the form (a, -a)
, where a
 ? 0.
When a = 1
, the point is 1, -1
.
When a = 2
, the point is 2, -2
.
When a = 3
, the point is 3, -3
.......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
You will get the equation of the line x + y = 0
.
Q u e s t i o n : 2 4
The linear equation 3x - 5y = has
a a unique solution
b two solutions
c infinitely many solutions
d no solution
S o l u t i o n :
c infinitely many solutions
Given linear equation: 3x - 5y = 15
Or, x = 
5y + 15
3
When y = 0
, x = 
15
3
 = 5
.
When  y = 3
, x = 
30
3
 = 10
.
When y = -3
, x = 
0
3
 = 0
.
Thus, we have the following table:
     x      5      10    0
     y      0      3    -3
Plot the points A(5, 0) , B(10, 3)
 and C(0, -3)
. Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation. 
Hence, the linear equation has infinitely many solutions.
Q u e s t i o n : 2 5
The equation 2x + 5y = 7 has a unique solution, if x and y are
a
natural numbers
b
rational numbers
c
positive real numbers
d
real numbers
S o l u t i o n :
Since, every point on the line represented by the equation 2x + 5y = 7 is its solution.
Therefore, there are infinite solutions of the equation the equation 2x + 5y = 7 in which the values of x and y are rational numbers,
positive real numbers or real numbers.
But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5y =
7.
So, the equation 2x + 5y = 7 has a unique solution, if x and y are both are natural numbers.
Hence, the correct option is a
.
Q u e s t i o n : 2 6
The graph of y = 5 is a line
a
making an intercept 5 on the x-axis
b
making an intercept 5 on the y-axis
c
parallel to the x-axis at a distance of 5 units from the origin
d
parallel to the y-axis at a distance of 5 units from the origin
S o l u t i o n :
As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.
?
The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.
So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.
Hence, the correct answers are options b
and c
.
D i s c l a i m e r : In this question, there are two correct answers.
Q u e s t i o n : 2 7
The graph of x = 4 is a line
a
making an intercept 4 on the x-axis
b
making an intercept 4 on the y-axis
c
parallel to the x-axis at a distance of 4 units from the origin
d
parallel to the y-axis at a distance of 4 units from the origin
S o l u t i o n :
As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.
?
 The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.
So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.
Hence, the correct options are a
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Q u e s t i o n : 2 0
The equation of the x-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the y-coordinate of any point on x-axis is always 0.
So, the equation of the x-axis is y = 0.
Hence, the correct option is b
.
Q u e s t i o n : 2 1
The equation of the y-axis is
a
  x = 0
b
y = 0
c
x = y
d
x + y = 0
S o l u t i o n :
Since, the x-coordinate of any point on y-axis is always 0.
So, the equation of the y-axis is x = 0.
Hence, the correct option is a
.
Q u e s t i o n : 2 2
The point of the form (a, a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
c
the line y = x
Given, a point of the form (a, a)
, where a
? 0 .
When a = 1
, the point is 1, 1
When a = 2
, the point is 2, 2
.....and so on.
Plot the points 1, 1
and 2, 2
.......and so on. Join the points and extend them in both the direction. You will get the equation of the line y = x
.
Q u e s t i o n : 2 3
The point of the form (a, -a), where a ? 0, lies on
a
the x-axis
b
the y-axis
c
the line y = x
d
the line x + y = 0
S o l u t i o n :
d
the line x + y = 0
Given, a point of the form (a, -a)
, where a
 ? 0.
When a = 1
, the point is 1, -1
.
When a = 2
, the point is 2, -2
.
When a = 3
, the point is 3, -3
.......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions. 
You will get the equation of the line x + y = 0
.
Q u e s t i o n : 2 4
The linear equation 3x - 5y = has
a a unique solution
b two solutions
c infinitely many solutions
d no solution
S o l u t i o n :
c infinitely many solutions
Given linear equation: 3x - 5y = 15
Or, x = 
5y + 15
3
When y = 0
, x = 
15
3
 = 5
.
When  y = 3
, x = 
30
3
 = 10
.
When y = -3
, x = 
0
3
 = 0
.
Thus, we have the following table:
     x      5      10    0
     y      0      3    -3
Plot the points A(5, 0) , B(10, 3)
 and C(0, -3)
. Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation. 
Hence, the linear equation has infinitely many solutions.
Q u e s t i o n : 2 5
The equation 2x + 5y = 7 has a unique solution, if x and y are
a
natural numbers
b
rational numbers
c
positive real numbers
d
real numbers
S o l u t i o n :
Since, every point on the line represented by the equation 2x + 5y = 7 is its solution.
Therefore, there are infinite solutions of the equation the equation 2x + 5y = 7 in which the values of x and y are rational numbers,
positive real numbers or real numbers.
But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5y =
7.
So, the equation 2x + 5y = 7 has a unique solution, if x and y are both are natural numbers.
Hence, the correct option is a
.
Q u e s t i o n : 2 6
The graph of y = 5 is a line
a
making an intercept 5 on the x-axis
b
making an intercept 5 on the y-axis
c
parallel to the x-axis at a distance of 5 units from the origin
d
parallel to the y-axis at a distance of 5 units from the origin
S o l u t i o n :
As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.
?
The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.
So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.
Hence, the correct answers are options b
and c
.
D i s c l a i m e r : In this question, there are two correct answers.
Q u e s t i o n : 2 7
The graph of x = 4 is a line
a
making an intercept 4 on the x-axis
b
making an intercept 4 on the y-axis
c
parallel to the x-axis at a distance of 4 units from the origin
d
parallel to the y-axis at a distance of 4 units from the origin
S o l u t i o n :
As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.
?
 The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.
So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.
Hence, the correct options are a
and d
.
D i s c l a i m e r : In this question, there are two correct answers.
Q u e s t i o n : 2 8
The graph of x + 3 = 0 is a line
a
making an intercept –3 on the x-axis
b
making an intercept –3 on the y-axis
c
parallel to the y-axis at a distance of 3 units to the left of y-axis
d
parallel to the x-axis at a distance of 3 units below the x-axis
S o l u t i o n :
As, the graph of x + 3 = 0 or x = -
3 is a line parallel to y-axis i.e. x = 0.
?
 The line represented by the equation x = -
3 is parallel to y-axis and intersects x-axis at x = -
3.
So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept -
3 on the x-axis.
Hence, the correct options are a
and c
.
D i s c l a i m e r : In this question, there are two correct answers.
Q u e s t i o n : 2 9
The graph of y + 2 = 0 is a line
a
making an intercept –2 on the x-axis
b
making an intercept –2 on the y-axis
c
parallel to the x-axis at a distance of 2 units below the x-axis
d
parallel to the y-axis at a distance of 2 units to the left of y-axis
S o l u t i o n :
As, the graph of y + 2 = 0 or y = -2 is a line parallel to x-axis i.e. y = 0.
? The line represented by the equation y = -2 is parallel to x-axis and intersects y-axis at y = -2.
So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept -2 on the y-axis.
Hence, the correct options are b
and c
.
D i s c l a i m e r : In this question, there are two correct answers.
Q u e s t i o n : 3 0
The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point
a
2, 0
b
3, 0
c